What is 2(3x+1)<= 20\land 4(x-1)<2 ?

The solution to 2(3x+1)<= 20\land 4(x-1)<2 is x< 3/2

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2(3x+1)<= 20\land 4(x-1)<2

2 (3 x + 1) \leq 20 and 4 (x - 1) < 2

x < \frac{3}{2}

Interval Notation

(- \infty, \frac{3}{2})

Decimal Notation

x < 1.5

Solution steps

2 (3 x + 1) \leq 20 and 4 (x - 1) < 2

2 (3 x + 1) \leq 20 : x \leq 3

4 (x - 1) < 2 : x < \frac{3}{2}

Combine the intervals

x \leq 3 and x < \frac{3}{2}

Merge Overlapping Intervals

x < \frac{3}{2}

\frac{m - 2}{m ^{2} - 5 m + 6}

19 - \frac{5}{2} x = 34

21 r - 3 = 3 (4 - 23 r)

6 x^{2} - 3 x + 12

3^{2} + 9^{2}

What is 2(3x+1)<= 20\land 4(x-1)<2 ?
The solution to 2(3x+1)<= 20\land 4(x-1)<2 is x< 3/2